ON SMALL RAMSEY NUMBERS IN GRAPHS
JANNE KORHONEN
[email protected]
Abstract. We give exact values for certain small 2-colour Ramsey numbers
in graphs. In particular, we prove that R(3, 3) = 6 and R(4, 4) = 18.
In the following, if G = (V, E) is a graph and U ⊆ V , we denote by G[U ] the
subgraph induced by U . For set X, we denote by [X]k the collection of all subsets
of X with k elements.
We will now prove some well-known results about certain small Ramsey numbers.
The following treatment of the matter along with the theorems and proofs is quite
standard; the results are originally from Greenwood and Gleason [1], except for
theorem 2, which is quite elementary and had appeared as an exercise in the William
Lowell Putnam Mathematical Competition held in March 1953. We use notation
from Radziszowski [2].
Definition 1. For k, l ≥ 2, denote by R(k, l) the smallest number N such that
for all graphs G = (V, E) with at least N vertices, G contains either a k-clique or
an independent set with l vertices. The values R(k, l) are called 2-colour Ramsey
numbers in graphs.
Now, we want to find out the exact values of R(k, l) for certain small values of k
and l. We start with an easy case.
Theorem 2. R(3, 3) = 6.
Proof. First, observe that the 5-cycle C5 does not contain a 3-clique or an independent set with 3 vertices. Thus, R(3, 3) > 5.
Now assume that G = (V, E) is a graph with |V | = 6. Let u ∈ V be an arbitrary
vertex. There are two possible scenarios:
(1) The set N = {v ∈ V | {u, v} ∈ E} has at least three elements. In this case,
either the set N is independent and the theorem holds, or we have two
adjacent vertices v1 , v2 ∈ N , in which case {u, v1 , v2 } is a clique and the
theorem also holds.
(2) The set {v ∈ V | {u, v} ∈ E} has at most two elements. Then by case (1),
there is a clique or a independent set of size 3 in the complement graph of
G and thus also in G.
In any case, we have that R(3, 3) ≤ 6.
To get tight bounds for R(4, 4), we need to see some more trouble. As a starting
point, we observe that R(m, 2) = R(2, m) = m for all m ≥ 2, because a graph with
m vertices is either Km or has two non-adjacent vertices, and on the other hand,
Km−1 serves as the proof for the lower bound.
This text was originally part of a course report for the course Deterministic Distributed
Algorithms (http://www.cs.helsinki.fi/u/josuomel/dda-2010/) at the University of Helsinki.
This has been separated as an independent text as it is the part of the report that might actually
be of interest for someone else.
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J. KORHONEN
Lemma 3. For all k, l ≥ 3, we have that
R(k, l) ≤ R(k − 1, l) + R(k, l − 1).
Proof. Let k, l ≥ 3. Let G = (V, E) be a graph with n = R(k − 1, l) + R(k, l − 1)
vertices. We show now that there is either a k-clique or an independent set with l
vertices in G.
Fix u ∈ V . Now we define
V+ (u) = {v ∈ V | {u, v} ∈ E}
V− (u) = {v ∈ V | {u, v} ∈
/ E}.
Observe that {u}, V+ (u) and V− (u) are disjoint and their union is V . Thus,
(1)
|V+ (u)| + |V− (u)| = R(k − 1, l) + R(k, l − 1) − 1.
This means that we must have |V+ (u)| ≥ R(k − 1, l) or |V− (u)| ≥ R(k, l − 1), because
otherwise inequality 1 would not hold. Thus, we now have two possible cases:
(1) We have |V+ (u)| ≥ R(k − 1, l), and therefore G[V+ (u)] either has a (k − 1)clique or an independent set with l vertices. In the latter case, we are done;
otherwise, there is K ⊆ V+ (u) such that K is a (k − 1)-clique. By the
definition of V+ (u), the set {u} ∪ K is a k-clique.
(2) We have |V− (u)| ≥ R(k, l − 1). Again, we either have a k-clique in G[V− (u)],
in which case the theorem holds, or then there is an independent set
I ⊆ V− (u) with |I| = l − 1. In the latter case {u} ∪ I is an independent set
with l vertices.
In some cases, the result of lemma 3 can be improved slightly.
Lemma 4. For all k, l ≥ 3, if R(k − 1, l) = 2p and R(k, l − 1) = 2q, then
R(k, l) ≤ R(k − 1, l) + R(k, l − 1) − 1.
Proof. Let k, l ≥ 3 such that R(k − 1, l) = 2p and R(k, l − 1) = 2q. Let G = (V, E)
be a graph with n = R(k − 1, l) + R(k, l − 1) − 1 = 2p + 2q − 1 vertices. Again, we
want to show that there is either a k-clique or an independent set with l vertices in
G.
Observe that if there is a vertex u ∈ V such that |V+ (u)| ≥ R(k − 1, l) or
|V− (u)| ≥ R(k − 1), then we can use same arguments as in the proof of lemma 3 to
see that there is a k-clique or an independent set with l vertices in G. Thus, it is
sufficient to show that such u exists.
The problematic case now that it might be that for all v ∈ V , |V+ (v)| =
R(k − 1, l) − 1 = 2p − 1 and |V− (v)| = R(k, l − 1) − 1 = 2q − 1. Assume that this is
in fact the case. In particular, then each vertex has degree 2p − 1, and thus there
are (2p − 1)(2p + 2q − 1)/2 edges in G. However, (2p − 1)(2p + 2q − 1)/2 is not an
integer, so this is not possible.
Lemma 5. R(4, 4) ≤ 18.
Proof. We have previously seen that
R(4, 2) = 4,
R(2, 4) = 4, and
R(3, 3) = 6.
ON SMALL RAMSEY NUMBERS IN GRAPHS
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Figure 1. Graph G.
Using lemmas 3 and 4, we get that
R(3, 4) ≤ R(2, 4) + R(3, 3) − 1 = 9,
R(4, 3) ≤ R(3, 3) + R(4, 2) − 1 = 9, and
R(4, 4) ≤ R(3, 4) + R(4, 3) = 18.
Lemma 6. R(4, 4) > 17.
Proof. We start by defining set
S17 = {x2 | x ∈ Z17 } \ {0} = {1, 2, 4, 8, 9, 13, 15, 16}.
Now let G = (Z17 , E), where
E = {{x, y} ∈ [Z17 ]2 | x − y ∈ S17 }.
(See figure 1.) Observe that since in the field Z17 it holds that −1 = 16 = 42 , if
x − y = a2 for some a, then y − x = (−1)(x − y) = (4a)2 , and thus G is well-defined.
Now suppose that K ⊆ Z17 is a 4-clique in G. We may in fact assume that 0 ∈ K,
because otherwise we get such a clique by subtracting the smallest element in K
from all the elements of K. Thus, suppose that K = {0, a, b, c}; by definition of G,
it holds that H = {a, b, c, a − b, a − c, b − c} ⊆ S17 . Since Z17 is a field, a−1 exists.
We define B = ba−1 and C = ca−1 ; these are distinct numbers and different from 1,
since a, b and c are distinct. Because a−1 = (n2 )−1 for some n ∈ Z17 , by multiplying
all elements of H by a−1 , we get that
{1, B, C, 1 − B, 1 − C, B − C} ⊆ S17 .
On the other hand, suppose that I ⊆ Z17 is an independent set in G with
4 elements. Again, we may assume that I = {0, a, b, c}. We have now that
J = {a, b, c, a − b, a − c, b − c} ⊆ Z17 \ (S17 ∪ {0}). It can be easily verified by testing
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J. KORHONEN
all possible cases that if x, y ∈ Z17 \ (S17 ∪ {0}), then xy ∈ S17 . Thus multiplying
all the elements of J by a−1 we see that
{1, B, C, 1 − B, 1 − C, B − C} ⊆ S17 ,
−1
where again B = ba and C = ca−1 .
We have seen that if there is a 4-clique or an independent set with 4 vertices in
G, then there are distinct number B, C ∈ S17 \ {1} such that
{1, B, C, 1 − B, 1 − C, B − C} ⊆ S17 .
We have that
1 − 2 = 16 ∈ S17
1 − 4 = 14 ∈
/ S17
1 − 8 = 10 ∈
/ S17
1 − 9 = 9 ∈ S17
1 − 13 = 5 ∈
/ S17
1 − 15 = 3 ∈
/ S17
1 − 16 = 1 ∈ S17 ,
and thus B, C ∈ {2, 9, 16}. However,
B = 9, C = 2 ⇒ B − C = 7 ∈
/ S17
B = 2, C = 9 ⇒ B − C = 10 ∈
/ S17
B = 16, C = 2 ⇒ B − C = 15 ∈
/ S17
B = 2, C = 16 ⇒ B − C = 3 ∈
/ S17
B = 16, C = 9 ⇒ B − C = 7 ∈
/ S17
B = 9, C = 16 ⇒ B − C = 10 ∈
/ S17 .
It follows that set {1, B, C, 1 − B, 1 − C, B − C} cannot be a subset of S17 . Thus, existence of 4-clique or an independent set with 4 vertices would lead to a contradiction
and is therefore not possible.
Since G is a graph with no 4-clique or independent set of 4 vertices, we have that
R(4, 4) > 17.
Combining the previous lemmas we get the following theorem.
Theorem 7. R(4, 4) = 18.
References
1. R.E. Greenwood and A.M. Gleason, Combinatorial relations and chromatic graphs, Canad. J.
Math 7 (1955), no. 1.
2. S.P. Radziszowski, Small Ramsey numbers, Electronic Journal of Combinatorics 1 (1994), last
updated 2009.